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Vì x là phân giác của D
=> D1 = D2 = D/2 = 90/2 = 45
Vì y là phân giác của B
=> B1 = B2 = B/2 = 90/2 = 45
Áp dụng tổng 3 góc của một tam giác , ta có :
B1 + H + C = 180
mà B1 = 45 ; C = 90
=> 45 + 90 + H = 180
=> H = 45
Vì H = D1
mà 2 góc nằm ở vị trí đồng vị
=> Dx//By ( điều phải chứng minh)
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1: Ta có:ABCD là hình chữ nhật
nên AB=CD;AD=BC
2: Xét tứ giác ABCD có
AB=CD
AD=BC
Do đó: ABCD là hình bình hành
Xét ΔADE và ΔCBF có
\(\widehat{D}=\widehat{B}\)
AD=CB
\(\widehat{DAE}=\widehat{BCF}\)
Do đó: ΔADE=ΔCBF
Suy ra: \(\widehat{AED}=\widehat{CFB}\)
=>\(\widehat{AEC}=\widehat{CFA}\)
Xét tứ giác AECF có
\(\widehat{AEC}=\widehat{CFA}\)
\(\widehat{FAE}=\widehat{FCE}\)
Do đó: AECF là hình bình hành
Suy ra: AE//CF
https://docs.google.com/document/d/1Wuo1vFdubrUg8F8-Ng_f-K8sda_JE_rRM704rtBrI-Q/edit?usp=sharing
Ta có H1+ H2+H3=180
E1+E2=180
mà E1=H1
nên E2=H2+H3
Tong 3 goc trong tam giác: E2+H2+A1=180
(H2+H3)+H2+A1=180
2.H2+H3+A1=180
SUY RA: H2=(180-90-A1):2 *** H3=90 hihi
=45-A1/2
mà A1=90-2A2
thay vào *** ta có H2=45-(90-2.A2)/2=A2
vậy H2=A2 hay EH//AD
Goi DE la phan giac cua goc D
Goi BF la phan giac cua goc B
Ta có góc B=D => B1=B2=D1=D2=B/2=D/2=90/2=45
Ta có D1=AED=45(so le trong)
B1=D1 =>AED=B1=45
ma AED=B1(o vi tri so le trong)
Suy ra DE//BF
Vậy tia phân giác góc B song song voi tia phan giac goc D